+1, interesting problem! My initial guess would be to do something like, take a transcendence basis $T$ for $\mathbb{R}$ over $\mathbb{Q}$, which will be uncountable, and take $S$ to be all points in the plane whose coordinates are in $T$. Not quite sure how to show this is dense, though.
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Zev ChonolesNov 29 '11 at 19:37

@Zev: Note that $T$ need not be dense, because for example it could be chosen to be contained in $(0,1)$, but since it is infinite its elements could be scaled by rationals to form another transcendence basis that is dense.
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Jonas MeyerNov 29 '11 at 20:05

1 Answer
1

1, 2, 3. Let $R$ be a cocountable subset of $\mathbb{R}$. We will construct a countable dense subset $S$ of $\mathbb{R}^2$ such that the ratio of any two distances between points in $S$ belongs to $R$. To do this, begin by placing two points $s_1, s_2 \in S$ unit distance apart. Now enumerate the disks with rational center and rational radius in $\mathbb{R}^2$ and place points $s_n$ in the interior of each such disk in turn satisfying the given condition. This is always possible because the set of all points at which $s_n$ cannot be placed is a countable union of sets of measure zero (one for each possible ratio of two distances lying outside of $S$), hence has measure zero.

The same argument together with transfinite induction should also establish 4, but I haven't thought about it too carefully.